c# - Convert Infix to Postfix with Binary Tree -

i'm trying make infix postfix converter. have searched code in google of algorithms stack or using lot's of regular expression or hard understand, want make binary tree.

here have did:

class node:

public class node {     private object element;     private node left;     private node right;     private node parent;      public node()     {     }      public node(object x, node r, node l)     {         element = x;         right = r;         left = l;     }      public object getelement()     {         return element;     }      public void setelement(object x)     {         element = x;     }      public node getleft()     {         return left;     }      public void setleft(node l)     {         left = l;     }      public node getright()     {         return right;     }      public void setright(node r)     {         right = r;     }      public node getparent()     {         return parent;     }      public void setparent(node p)     {         parent = p;     } } 

sorry using get/set instead of simple auto property.

and bst class has insert, postfix, prefix , infix method (i have writen search, deletemin, etc don't need them question)

bst class:

public class binarysearchtree {     //node r: root.     public void insert(node r, node p, object x)     {         if (r == null)         {             r = new node(x, null, null);             r.setparent(p);             if ((int)x < (int)p.getelement())                 p.setleft(r);             else if ((int)x > (int)p.getelement())                 p.setright(r);         }         if ((int) x < (int) r.getelement())             insert(r.getleft(), r, x);         else if ((int) x > (int) r.getelement())             insert(r.getright(), r, x);     }      public void preorder(node p)     {         if (p != null)         {             console.writeline(p.getelement());             preorder(p.getleft());             preorder(p.getright());         }     }      public void inorder(node p)     {         if (p != null)         {             inorder(p.getleft());             console.writeline(p.getelement());             inorder(p.getright());         }     }      public void postorder(node p)     {         if (p != null)         {             postorder(p.getleft());             postorder(p.getright());             console.writeline(p.getelement());         }     } } 

my insert method work numbers example:
if want write inorder, preorder , post order of below tree

preorder: 5, 3, 2, 4, 7, 6, 12
postorder: 2, 4, 3, 6, 12, 7, 5
inorder: 2, 3, 4, 5, 6, 7, 12

main method example:

private static void main() {     binarysearchtree bst = new binarysearchtree();     node r = new node(5, null, null);     bst.insert(r, null, 3);     bst.insert(r, null, 7);     bst.insert(r, null, 4);     bst.insert(r, null, 12);     bst.insert(r, null, 2);     bst.insert(r, null, 6);     bst.inorder(r);     console.writeline();     bst.postorder(r);     console.writeline();     bst.preorder(r); } 

now want make infix postfix converter don't know how implement insert method that, , how handle operator , operands in order. , problem parentheses.
grateful if me code.

an example tree:

mathblog has infix postfix converter, can check yours it.
http://www.mathblog.dk/tools/infix-postfix-converter/

it looks easiest way convert expression infix postfix notation use standard stack based algorithm(it has linear time complexity, optimal) , build tree(constructing tree postfix expression simple because operators in correct order).